Answer
2.2 m
Work Step by Step
$\vec{A}=A\hat{i},$ (directed along the x axis)
$\vec{B}=B_{x}\hat{i}+B_{y}\hat{j}$
$ \sqrt{B_{x}^{2}+B_{y}^{2}}=7.0$ m.
First,
$\vec{A}+\vec{B}=(A+B_{x}\hat{)i}+B_{y}\hat{j}=B_{y}\hat{j}$
$A+B_{x}=0$ because the sum is directed along the y axis.
Magnitude is 3.0A, so $B_{y}=3.0A$.
So $A^{2}=B_{x}^{2}$,
and
$\sqrt{A^{2}+B_{y}^{2}}=7.0m$
Since we know that:
$A^{2}+(3.0A)^{2}=49m^{2}$
$10A^{2}=49m^{2}$
$A=\displaystyle \frac{7.0m}{\sqrt{10}}=2.2$ m
